---
title: "Repeatability assumption, select best of three projects"
subject: "Engineering Economics"
chapter: 4
key: "semester_6/engineering_economics:repeatability-abc-14"
source: https://www.bctengineer.com/q/semester_6/engineering_economics/repeatability-abc-14
site: BCT Engineering
---

# Repeatability assumption, select best of three projects

Engineering Economics · Chapter 4: Comparative Analysis of Alternatives

- Asked in: 2076 Shrawan, 6 marks; 2071 Bhadra, 6 marks (asked 2 times)
- Video: [Repeatability assumption |Economics |Board Exam Solution in NEPALI](https://www.youtube.com/watch?v=OfLkWltuILI) — OPengineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/engineering_economics/repeatability-abc-14
- Whole chapter: https://www.bctengineer.com/md/semester_6/engineering_economics/chapter-4

## Question

Use repeatability assumption to select the best project from the following three projects.
| Project | A | B | C |
|---|---|---|---|
| Initial Investment | 100000 | 200000 | 250000 |
| Annual Expenditure | 25000 | 20000 | 15000 |
| Useful Life, years | 3 | 5 | 7 |
| Salvage Value | 40000 | 50000 | 60000 |
| MARR | 14% | 14% | 14% |

## Short answer

Project A selected with lowest equivalent annual cost Rs 56,444/year at 14% MARR using repeatability assumption. AW calculated for each project's life cycle: A (3 yr), B (5 yr), C (7 yr). All have only costs, so minimum AW wins.

## Answer

Repeatability assumption: each project is repeated, with the same cash flows, as many times as needed, so all projects can be compared over a common (very long) period. Equal annual worth (AW) over one life cycle then represents every cycle, and lives need not be equal. All three projects have only costs, so the best is the one with the smallest annual cost (largest, i.e. least negative, AW).

AW = -I(A/P, i, N) - annual expenditure + S(A/F, i, N), MARR = 14%.

**Project A** (N = 3): (A/P,14%,3) = 0.4307, (A/F,14%,3) = 0.2907
AW = -100,000(0.4307) - 25,000 + 40,000(0.2907) = -43,073 - 25,000 + 11,629 = **-56,444**

**Project B** (N = 5): (A/P,14%,5) = 0.2913, (A/F,14%,5) = 0.1513
AW = -200,000(0.2913) - 20,000 + 50,000(0.1513) = -58,257 - 20,000 + 7,564 = **-70,693**

**Project C** (N = 7): (A/P,14%,7) = 0.2332, (A/F,14%,7) = 0.0932
AW = -250,000(0.2332) - 15,000 + 60,000(0.0932) = -58,298 - 15,000 + 5,592 = **-67,707**

| Project | Life | AW (Rs./year) |
|---|---|---|
| A | 3 | -56,444 |
| B | 5 | -70,693 |
| C | 7 | -67,707 |

**Answer: Select Project A**, the one with the lowest equivalent annual cost (Rs. 56,444 per year). B and C are costlier by about Rs. 14,249 and Rs. 11,263 per year respectively. (No revenue is given, so the projects are compared on cost alone, assuming they give equal service.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What does the repeatability assumption say?
   A. Lives must be equal
   B. Projects are used once only
   C. Salvage is reinvested
   D. Projects are repeated with same cash flows, so one cycle's AW represents all
   Answer: D. With identical repeats, comparing AW over one cycle is valid even for unequal lives. (Key point: Repeatability assumption)

2. Which AW formula is used for these cost-only projects?
   A. -I(A/F) + S(A/P)
   B. -I(A/P) - annual cost + S(A/F)
   C. -I(P/A) + S(P/F)
   D. I(A/P) + annual cost + S(A/F)
   Answer: B. Capital recovery is -I(A/P), less annual cost, plus salvage S(A/F). (Key point: AW formula)

3. Which project is chosen and why?
   A. B, lowest annual expenditure
   B. B, largest investment
   C. C, longest life
   D. A, smallest annual cost (AW -56,444)
   Answer: D. A has the least negative AW; B (-70,693) and C (-67,707) cost more each year. (Key point: Selection rule)

## Related in Engineering Economics

- [IRR to select best of A, B, C (15 years)](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Airr-select-abc-15yr) — Engineering Economics, chapter 4
- [Repeatability, 4-year study, infinite life](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aselect-best-repeatability-study-infinite) — Engineering Economics, chapter 4
- [Best of three mutually exclusive projects, 10-year study](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Amutually-exclusive-abc-study-10-years) — Engineering Economics, chapter 4
- [Repeatability comparison of projects A and B](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-a-b-200000-300000) — Engineering Economics, chapter 4
- [Repeatability, best of A and B](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Arepeatability-a-b-400000-700000) — Engineering Economics, chapter 4
- [Incremental IRR to select best of A to D](https://www.bctengineer.com/question.md?key=semester_6%2Fengineering_economics%3Aincremental-irr-abcd-salvage-25) — Engineering Economics, chapter 4

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